Question

A series circuit contains a 3.00 H inductor, a 3.00

Answer 1

The series circuit contains L = 3.00 H,C = 3.00 ?F = 3.00* 10^-6 F and a resistor R = 30.0 ?

The series circuit is connected to a 120 V (rms) source ofvariable frequency.

The total impedance of the circuit is

Z = (R² + (X_L -X_C)²)^1/2--------------------(1)

Here,X_L = wL = 2?fL and X_C = (1/wC)= (1/2?fC)

Here,f = (1/2?) * (1/LC)^1/2

or f = (1/2 * 3.14) * (1/3.00 * 3.00 *10^-6)^1/2

or f = 53.0 Hz

Here,X_L = 2 * 3.14 * 53.0 * 3.00 = 998.52?

and X_C = (1/2 * 3.14 * 53.0 * 3.00 *10^-6) = 1001.5 ?

Substituting the values in equation (1),we get

Z = ((30.0)² + (998.52 -1001.5)²)^1/2

or Z = 30.1 ?

The current through the circuit is

I_rms = (V_rms/Z)

or I_rms = (120/30.1) = 3.9 A

(a)Therefore,the power delivered to the circuit when thefrequency of the source is the resonance frequency is

P = I_rms² * Z = (3.9)² * 30.1= 457.8 W

(b)When the frequency is one-half the resonancefrequency,then

X_L = 2? * (f/2) * L = ?fL = 3.14 * 53.0 *3.00 = 499.26 ?

and X_C = (1/2?fC) = (1/2? * (f/2) * C) =(1/?fC) = (1/3.14 * 53.0 * 3.00 * 10^-6) = 500.75?

Substituting the values in equation (1),we get

Z₁ = ((30.0)² + (499.26 -500.75)²)^1/2

or Z₁ = 15.05 ?

Therefore,the power delivered to the circuit when thefrequency of the source is one-half the resonance frequencyis

P = (3.9)² * 15.05 = 228.9 W

(c)The impedance of the circuit when the frequency of thesource is one-fourth the resonance frequency is

Z₂ = (Z/4) = (30.1/4) = 7.525 ?

Therefore,the power delivered to the circuit when thefrequency of the source is one-fourth the resonance frequencyis

P = (3.9)² * 7.525 = 114.4 W

(d)The impedance of the circuit when the frequency of thesource is two times the resonance frequency is

Z₂ = 2Z = 2 * 30.1 = 60.2 ?

Therefore,the power delivered to the circuit when thefrequency of the source is two times the resonance frequencyis

P = (3.9)² * 60.2 = 915.6 W

(e)The impedance of the circuit when the frequency of thesource is four times the resonance frequency is

Z₂ = 4Z = 4 * 30.1 = 120.4 ?

Therefore,the power delivered to the circuit when thefrequency of the source is two times the resonance frequencyis

P = (3.9)² * 120.4 = 1831.3 W

The maximum power is delivered to the circuit when thefrequency of the source is four times the resonance frequency.Thisis because the power delivered to the circuit depends directly onthe impedance of the circuit.